An arc lamp requires a direct current of \(10\ A\) at \(80\ V\) to function. If it is connected to a \( 220\ V (rms ), 50\ Hz\ AC\) supply, the series inductor needed for it to work is close to :

\(\mathrm{i}=\frac{\mathrm{e}}{\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}}}=\frac{\mathrm{e}}{\sqrt{\mathrm{R}^{2}+\omega^{2} \mathrm{L}^{2}}}=\frac{\mathrm{e}}{\sqrt{\mathrm{R}^{2}+4 \pi^{2} \mathrm{v}^{2} \mathrm{L}^{2}}}\)

\(10=\frac{220}{\sqrt{64+4 \pi^{2}(50)^{2} \mathrm{L}}}\left[\because \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{80}{10}=8\right]\)

On solving we get

\(\mathrm{L}=0.065\, \mathrm{H}\)